Theorem nss 3986

Description: Negation of subclass relationship. Exercise 13 of [TakeutiZaring] p. 18. (Contributed by NM, 25-Feb-1996.) (Proof shortened by Andrew Salmon, 21-Jun-2011.)

Assertion

Ref	Expression
nss	⊢ (¬ 𝐴 ⊆ 𝐵 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Proof of Theorem nss

Step	Hyp	Ref	Expression
1		exanali 1866	. . 3 ⊢ (∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) ↔ ¬ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
2		df-ss 3907	. . 3 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
3	1, 2	xchbinxr 336	. 2 ⊢ (∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) ↔ ¬ 𝐴 ⊆ 𝐵)
4	3	bicomi 225	1 ⊢ (¬ 𝐴 ⊆ 𝐵 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 207   ∧ wa 396  ∀wal 1545  ∃wex 1786   ∈ wcel 2119   ⊆ wss 3890

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816

This theorem depends on definitions:  df-bi 208  df-an 397  df-ex 1787  df-ss 3907

This theorem is referenced by:  grur1  10741  psslinpr  10952  reclem2pr  10969  mreexexlem2d  17609  prmcyg  19867  filconn  23873  alexsubALTlem4  24040  wilthlem2  27057  shne0i  31544  onvf1odlem2  35339  erdszelem10  35435  fundmpss  36002  ntrneineine1lem  44535  nssrex  45540  nssd  45559  nsssmfmbf  47229